Problem: Simplify and expand the following expression: $ \dfrac{z - 9}{z + 4}-\dfrac{4z + 9}{z + 3} $
Answer: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(z + 4)(z + 3)$ Multiply the first term by $\dfrac{z + 3}{z + 3}$ $ \begin{align*} \dfrac{z - 9}{z + 4} \times \dfrac{z + 3}{z + 3} & = \dfrac{(z - 9)(z + 3)}{(z + 4)(z + 3)} \\ & = \dfrac{z^2 - 6z - 27}{(z + 4)(z + 3)}\end{align*} $ Multiply the second term by $\dfrac{z + 4}{z + 4}$ $ \begin{align*} \dfrac{4z + 9}{z + 3} \times \dfrac{z + 4}{z + 4} & = \dfrac{(4z + 9)(z + 4)}{(z + 3)(z + 4)} \\ & = \dfrac{4z^2 + 25z + 36}{(z + 3)(z + 4)}\end{align*} $ Now we have: $ = \dfrac{z^2 - 6z - 27}{(z + 4)(z + 3)} - \dfrac{4z^2 + 25z + 36}{(z + 3)(z + 4)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{z^2 - 6z - 27 - (4z^2 + 25z + 36)}{(z + 4)(z + 3)} $ $ = \dfrac{z^2 - 6z - 27 - 4z^2 - 25z - 36}{(z + 4)(z + 3)} $ $ = \dfrac{-3z^2 - 31z - 63}{(z + 4)(z + 3)}$ Expand the denominator: $ = \dfrac{-3z^2 - 31z - 63}{z^2 + 7z + 12}$